Proof of Theorem clsun
Step | Hyp | Ref
| Expression |
1 | | difundi 3879 |
. . . . . 6
⊢ (𝑋 ∖ (𝐴 ∪ 𝐵)) = ((𝑋 ∖ 𝐴) ∩ (𝑋 ∖ 𝐵)) |
2 | 1 | fveq2i 6194 |
. . . . 5
⊢
((int‘𝐽)‘(𝑋 ∖ (𝐴 ∪ 𝐵))) = ((int‘𝐽)‘((𝑋 ∖ 𝐴) ∩ (𝑋 ∖ 𝐵))) |
3 | | difss 3737 |
. . . . . . 7
⊢ (𝑋 ∖ 𝐴) ⊆ 𝑋 |
4 | | difss 3737 |
. . . . . . 7
⊢ (𝑋 ∖ 𝐵) ⊆ 𝑋 |
5 | | clsun.1 |
. . . . . . . 8
⊢ 𝑋 = ∪
𝐽 |
6 | 5 | ntrin 20865 |
. . . . . . 7
⊢ ((𝐽 ∈ Top ∧ (𝑋 ∖ 𝐴) ⊆ 𝑋 ∧ (𝑋 ∖ 𝐵) ⊆ 𝑋) → ((int‘𝐽)‘((𝑋 ∖ 𝐴) ∩ (𝑋 ∖ 𝐵))) = (((int‘𝐽)‘(𝑋 ∖ 𝐴)) ∩ ((int‘𝐽)‘(𝑋 ∖ 𝐵)))) |
7 | 3, 4, 6 | mp3an23 1416 |
. . . . . 6
⊢ (𝐽 ∈ Top →
((int‘𝐽)‘((𝑋 ∖ 𝐴) ∩ (𝑋 ∖ 𝐵))) = (((int‘𝐽)‘(𝑋 ∖ 𝐴)) ∩ ((int‘𝐽)‘(𝑋 ∖ 𝐵)))) |
8 | 7 | 3ad2ant1 1082 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘((𝑋 ∖ 𝐴) ∩ (𝑋 ∖ 𝐵))) = (((int‘𝐽)‘(𝑋 ∖ 𝐴)) ∩ ((int‘𝐽)‘(𝑋 ∖ 𝐵)))) |
9 | 2, 8 | syl5eq 2668 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝑋 ∖ (𝐴 ∪ 𝐵))) = (((int‘𝐽)‘(𝑋 ∖ 𝐴)) ∩ ((int‘𝐽)‘(𝑋 ∖ 𝐵)))) |
10 | | simp1 1061 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → 𝐽 ∈ Top) |
11 | | unss 3787 |
. . . . . . 7
⊢ ((𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) ↔ (𝐴 ∪ 𝐵) ⊆ 𝑋) |
12 | 11 | biimpi 206 |
. . . . . 6
⊢ ((𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (𝐴 ∪ 𝐵) ⊆ 𝑋) |
13 | 12 | 3adant1 1079 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (𝐴 ∪ 𝐵) ⊆ 𝑋) |
14 | 5 | ntrdif 20856 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ (𝐴 ∪ 𝐵) ⊆ 𝑋) → ((int‘𝐽)‘(𝑋 ∖ (𝐴 ∪ 𝐵))) = (𝑋 ∖ ((cls‘𝐽)‘(𝐴 ∪ 𝐵)))) |
15 | 10, 13, 14 | syl2anc 693 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝑋 ∖ (𝐴 ∪ 𝐵))) = (𝑋 ∖ ((cls‘𝐽)‘(𝐴 ∪ 𝐵)))) |
16 | 5 | ntrdif 20856 |
. . . . . . 7
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋) → ((int‘𝐽)‘(𝑋 ∖ 𝐴)) = (𝑋 ∖ ((cls‘𝐽)‘𝐴))) |
17 | 16 | 3adant3 1081 |
. . . . . 6
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝑋 ∖ 𝐴)) = (𝑋 ∖ ((cls‘𝐽)‘𝐴))) |
18 | 5 | ntrdif 20856 |
. . . . . . 7
⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝑋 ∖ 𝐵)) = (𝑋 ∖ ((cls‘𝐽)‘𝐵))) |
19 | 18 | 3adant2 1080 |
. . . . . 6
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝑋 ∖ 𝐵)) = (𝑋 ∖ ((cls‘𝐽)‘𝐵))) |
20 | 17, 19 | ineq12d 3815 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (((int‘𝐽)‘(𝑋 ∖ 𝐴)) ∩ ((int‘𝐽)‘(𝑋 ∖ 𝐵))) = ((𝑋 ∖ ((cls‘𝐽)‘𝐴)) ∩ (𝑋 ∖ ((cls‘𝐽)‘𝐵)))) |
21 | | difundi 3879 |
. . . . 5
⊢ (𝑋 ∖ (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵))) = ((𝑋 ∖ ((cls‘𝐽)‘𝐴)) ∩ (𝑋 ∖ ((cls‘𝐽)‘𝐵))) |
22 | 20, 21 | syl6eqr 2674 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (((int‘𝐽)‘(𝑋 ∖ 𝐴)) ∩ ((int‘𝐽)‘(𝑋 ∖ 𝐵))) = (𝑋 ∖ (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵)))) |
23 | 9, 15, 22 | 3eqtr3d 2664 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (𝑋 ∖ ((cls‘𝐽)‘(𝐴 ∪ 𝐵))) = (𝑋 ∖ (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵)))) |
24 | 23 | difeq2d 3728 |
. 2
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (𝑋 ∖ (𝑋 ∖ ((cls‘𝐽)‘(𝐴 ∪ 𝐵)))) = (𝑋 ∖ (𝑋 ∖ (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵))))) |
25 | 5 | clscld 20851 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ (𝐴 ∪ 𝐵) ⊆ 𝑋) → ((cls‘𝐽)‘(𝐴 ∪ 𝐵)) ∈ (Clsd‘𝐽)) |
26 | 10, 13, 25 | syl2anc 693 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((cls‘𝐽)‘(𝐴 ∪ 𝐵)) ∈ (Clsd‘𝐽)) |
27 | 5 | cldss 20833 |
. . . 4
⊢
(((cls‘𝐽)‘(𝐴 ∪ 𝐵)) ∈ (Clsd‘𝐽) → ((cls‘𝐽)‘(𝐴 ∪ 𝐵)) ⊆ 𝑋) |
28 | 26, 27 | syl 17 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((cls‘𝐽)‘(𝐴 ∪ 𝐵)) ⊆ 𝑋) |
29 | | dfss4 3858 |
. . 3
⊢
(((cls‘𝐽)‘(𝐴 ∪ 𝐵)) ⊆ 𝑋 ↔ (𝑋 ∖ (𝑋 ∖ ((cls‘𝐽)‘(𝐴 ∪ 𝐵)))) = ((cls‘𝐽)‘(𝐴 ∪ 𝐵))) |
30 | 28, 29 | sylib 208 |
. 2
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (𝑋 ∖ (𝑋 ∖ ((cls‘𝐽)‘(𝐴 ∪ 𝐵)))) = ((cls‘𝐽)‘(𝐴 ∪ 𝐵))) |
31 | 5 | clsss3 20863 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋) → ((cls‘𝐽)‘𝐴) ⊆ 𝑋) |
32 | 31 | 3adant3 1081 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((cls‘𝐽)‘𝐴) ⊆ 𝑋) |
33 | 5 | clsss3 20863 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝑋) → ((cls‘𝐽)‘𝐵) ⊆ 𝑋) |
34 | 33 | 3adant2 1080 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((cls‘𝐽)‘𝐵) ⊆ 𝑋) |
35 | 32, 34 | jca 554 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (((cls‘𝐽)‘𝐴) ⊆ 𝑋 ∧ ((cls‘𝐽)‘𝐵) ⊆ 𝑋)) |
36 | | unss 3787 |
. . . 4
⊢
((((cls‘𝐽)‘𝐴) ⊆ 𝑋 ∧ ((cls‘𝐽)‘𝐵) ⊆ 𝑋) ↔ (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵)) ⊆ 𝑋) |
37 | | dfss4 3858 |
. . . 4
⊢
((((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵)) ⊆ 𝑋 ↔ (𝑋 ∖ (𝑋 ∖ (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵)))) = (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵))) |
38 | 36, 37 | bitri 264 |
. . 3
⊢
((((cls‘𝐽)‘𝐴) ⊆ 𝑋 ∧ ((cls‘𝐽)‘𝐵) ⊆ 𝑋) ↔ (𝑋 ∖ (𝑋 ∖ (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵)))) = (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵))) |
39 | 35, 38 | sylib 208 |
. 2
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (𝑋 ∖ (𝑋 ∖ (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵)))) = (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵))) |
40 | 24, 30, 39 | 3eqtr3d 2664 |
1
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((cls‘𝐽)‘(𝐴 ∪ 𝐵)) = (((cls‘𝐽)‘𝐴) ∪ ((cls‘𝐽)‘𝐵))) |